Squares Galore

Let $P(x)= x_1(x+1)^2+x_2(x+2)^2+\cdots+x_7(x+7)^2$ Note that $P(x)$ is a quadratic polynomial. Then the given conditions are simply $P(0)=1$ , $P(1)=10$ , and $P(3)=100$ . Then by the Lagrange Interpolation Formula $P(2)=1\cdot \frac{(2-1)(2-3)}{(0-1)(0-3)}+10\cdot\frac{(2-0)(2-3)}{(1-0)(1-3)}+100\cdot \frac{(2-0)(2-1)}{(3-0)(3-1)}=\boxed{43}$

The key here is to flip your idea of what you call the variables. Instead of seeing x as the variable, consider the coefficients. For example, we can rewrite the first equation as:

$(1)^2(x1) + 2^2 (x2) + .... + 7^2 (x7) = 1.$ This is more striking if we write expand the coefficients as functions of the first term:

$(1)^2 (x1) + (1+1)^2 (x2) + (1+2) x^3 +..... + (1+6)^2 (x7) = 1.$

Likewise, the other two can be rewritten with as

$(2)^2 (x1) + (2+1)^2 (x2) + (2+2) (x3) + ... + (2+6)^2 (x7) = 10,$

$(4)^2 (x1) + (4+1)^2(x2) + (4+3)^2(x3) + ..... + (4+6)^2(x7) = 100.$

Now, let $x_1, x_2, \ldots x_7$ be fixed, and consider the polynomial

$(i)^2 (x1) + (i+1)^2 (x2) + ....... + (i+6)^2(x7) = f(i)$ . Because the left side is a quadratic in $i$ (with constant coefficients), and we are given three values of the quadratic, by the identity theorem the polynomial is uniquely determined. Thus, the right side is of degree two, and we can easily find $f(i) = 12i^2 - 27i + 16$ . Thus, our general polynomial, true for any value of i, is $(i)^2 (x1) + (i+1)^2 (x2) + ....... + (i+6)^2(x7) = 12x^2 - 27x + 16$ . The value we desire is seen to be $i=3$ , so plugging this in to the right side, we get $12(9) - 27(3) + 16 = 43$

1+3+\ldots+ $2k-1$ = k^2 Let z be sought value. Eq3 - Eq1 yields 3\cdot $5x_1 +7x_2 +\ldots + 17 x_7$ = 99 hence \cdot $5x_1 +7x_2 +\ldots + 17 x_7$ =33 Since Eq2 + \cdot $5x_1 +7x_2 +\ldots + 17 x_7$ = z we obtain 10 + 33 = z

• $\displaystyle \sum_{m=1}^7 m^2 x_{m}=1$ ;
• $\displaystyle \sum_{m=1}^7 (m+1)^2 x_{m}=10$ ;
• $\displaystyle \sum_{m=1}^7 (m+3)^2 x_{m}=100$ ;
• $\displaystyle \sum_{m=1}^7 (m+2)^2 x_{m}=?$
• What we have to solve is:
• $p \displaystyle \sum_{m=1}^7 m^2 x_{m} + q \displaystyle \sum_{m=1}^7 (m+1)^2 x_{m} + r \displaystyle \sum_{m=1}^7 (m+3)^2 x_{m} = s \displaystyle \sum_{m=1}^7 (m+2)^2 x_{m}$ .........(1)
• $\Leftrightarrow$ $\displaystyle \sum_{m=1}^7 pm^2 x_{m} + \displaystyle \sum_{m=1}^7 q(m+1)^2 x_{m} + \displaystyle \sum_{m=1}^7 r(m+3)^2 x_{m} = \displaystyle \sum_{m=1}^7 s(m+2)^2 x_{m}$
• $\Leftrightarrow$ $pm^2 x_{m} + q(m+1)^2 x_{m} + r(m+3)^2 x_{m} = s(m+2)^2 x_{m}$
• $\Leftrightarrow$ $pm^2 + q(m+1)^2 + r(m+3)^2 = s(m+2)^2$
• $\Leftrightarrow$ $pm^2 + q(m^2+2m+1) + r(m^2+6m+9) = s(m^2+4m+4)$
• $\Leftrightarrow$ $(p+q+r)m^2+(2q+6r)m+(q+9r)=sm^2+4sm+4s$ ...........(2)
• According to Right Hand Side,
• $(p+q+r):(2q+6r):(q+9r)=1:4:4$ ........(3)
• So, $2q+6r=q+9r$
• $\Leftrightarrow q=3r$
• $\Leftrightarrow \frac {q}{r}= \frac {3}{1}$
• $\Leftrightarrow q:r=3:1$
• Let us take $q=3k$ and $r=k$
• Now, From (3),
• $(p+q+r):(2q+6r)=1:4$
• $\Leftrightarrow (p+4k):(12k)=1:4$
• $\Leftrightarrow 4(p+4k)=12k$
• $\Leftrightarrow 4p=-4k$
• $\Leftrightarrow p=-k$
• From 1st terms on both sides of (2), we get,
• $p+q+r=s$
• $3k=s$
• Putting the values of p, q, r & the summation series in (1) simultaneously, we get,
• $-k*1+3k*10+k*100=3k*?$
• $-k+30k+100k=3k?$
• $3k?=129k$
• $?=43$
• [This is the formal solution. Modification permitted.]
• Rabeeb Ibrat, Class Ten, Cantonment Public School & College, Mymensingh, Bangladesh.

(16x1- x1)/3 + 4x1= 9x1 (25x2- 4x2) + 9x2 = 16 x2 and so,

[(16x1+25x2+36x3+49x4+64x5+81x6+100x7)-(x1+4x2+9x3+16x4+25x5+36x6+49x7)]3 + 4x1+9x2+16x3+25x4+36x5+49x6+64x7 = 9x1+16x2+25x3+36x4+49x5+64x6+81x7

(100-1)/3 +10= 43

Since there are $7$ variables but only $3$ equations, we suspect that we can write the desired value as a sum of multiples of the given equations.

After some experimentation, we find that

$\left(-\frac{1}{3}\right)(x_1+4x+2+9x_3+16x_4+25x_5+36x_6+49x_7)$ $+(1)(4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7)$ $+\left(\frac{1}{3}\right)(16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7)$ $=9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7$

The desired value is therefore $\left(-\frac{1}{3}\right)(1)+(1)(10)+\left(\frac{1}{3}\right)(100)=43$ .

Some motivation for these coefficients in the linear combination can be taken from ignoring the last $4$ terms and solving the resulting three-variable system of equations whose variables are the coefficients.

Let $S_1 = x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7$ , $S_2 = 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7$ , $S_3 = 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 54x_6 + 81x_7$ and $S_4 = 16x_1 + 25x_2 + 36x_3 + 49x_4 + 64x_5 + 81x_6 + 100x_7$ .

We notice that each of the coefficients of $x_i$ in $S_i$ are $i^2$ , $(i+1)^2$ , $(i+2)^2$ and $(i+3)^2$ respectively.

Solution 1: Let Now, let $x_1, x_2, \ldots x_7$ be fixed, and consider the polynomial

$f(n) = (n)^2 (x_1) + (n+1)^2 (x_2) + \ldots + (n+6)^2(x_7)$ . Because the left side is a quadratic in $n$ (with constant coefficients), and we are given three values of the quadratic, Hence the quadratic must be $f(n) = 12n^2 - 27n + 16$ . Thus, the answer is given by $n=3$ , $f(n) = 12(9) - 27(3) + 16 = 43$

Solution 2: Thus, we seek a linear combination such that: $ai^2 + b(i+1)^2 + c(i+3)^2 = (i+2)^2 \Rightarrow (a + b +c)i^2 + (2b + 6c)i + b + 9c = i^2 + 4i + 4$ . Equating coefficients, we have $a + b + c = 1$ , $2b + 6c = 4 \Rightarrow b + 3c = 2$ and $b + 9c = 4$ . Subtracting the last two equations gives $c = \frac{1}{3}$ and $b = 1$ . Substituting these values in the first equation we have $a = -\frac{1}{3}$ .

Therefore $S_3 = -\frac{1}{3}S_1 + S_2 + \frac{1}{3}S_4 = -\frac{1}{3}\cdot 1 + 1 \cdot 10 + \frac{1}{3}\cdot 100 = 43$ .